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Approximation Algorithms for TSP Tsvi Kopelowitz Ariel Rosenfeld 1

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HC-Hamiltonian Cycle Input: graph G=(V,E) Output: a cycle tour in G that visits each vertex exactly once. Problem is known to be NP-Hard. 2

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TSP – Traveling Salesman Problem Input: a complete graph G=(V,E) with edges of non-negative cost (c(e)). Output: find a cycle tour of minimum cost that visits each vertex exactly once. The problem is NP-Hard (reduction from HC). … and hard to approximate. 3

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Hardness of Approximation Claim: For every c>1, there is no polynomial time algorithm which can approximate TSP within a factor of c, unless P=NP. 4

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Proof By reduction from Hamiltonian cycle. ◦Given a graph G, we want to determine if it has a HC. ◦Construct a complete graph G’ with same vertices as G, where each edge e has weight 1 if it is in G, and weight c*n if it is not (n is the number of vertices). 5

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Example 6 G=

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Example c*n 7 G’=

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Proof Run algorithm A for solving TSP within a factor of c. ◦If there is a HC, TSP has a solution of weight n, and approximation is at most c*n ◦If there is no HC, then every tour in TSP has at least one edge of weight c*n, so weight of tour is at least c*n +n-1. 8

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Metric A (complete) graph with weight function w on the edges is called a metric if: ◦For any two vertices u,v in the graph: w(u,v)=w(v,u) ◦The Triangle Equality holds in the graph. 9

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Triangle Inequality Recall: The triangle inequality holds in a (complete) graph with weight function w on the edges if for any three vertices u,v,x in the graph: Metric TSP is still NP-hard, but now we can approximate. 10

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2-approximation Given G, construct an MST T for G ◦(since it is a metric graph, it doesn’t matter whether it is directed or not). For each edge in T create a double edge. (This will guarantee that the degrees of all the vertices are even) Find an Euler tour in the doubled T. 11

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2-approximation Given G, construct an MST T for G ◦(since it is a metric graph, it doesn’t matter whether it is directed or not). For each edge in T create a double edge. (This will guarantee that the degrees of all the vertices are even) Find an Euler tour in the doubled T. Create shortcuts in the Euler tour, to create a tour. 13

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2-approximation How will we create these “shortcuts”? ◦Traverse the Euler tour. ◦Whenever the Euler tour returns to a vertex already visited, “skip” that vertex. The process creates a Hamiltonian cycle. 14

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2-approximation Claim: The above algorithm gives a 2- approximation for the TSP problem (in a metric graph with the triangle inequality). Proof: Definitions: OPT is the optimal solution (set of edges) for TSP. A is the set of edges chosen by the algorithm. EC is the set of edges in the Euler cycle. 17

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2-approximation Proof Continued: cost(T) cost(OPT): ◦since OPT is a cycle, remove any edge and obtain a spanning tree. cost(EC) = 2cost(T) 2cost(OPT). cost(A) cost(EC) ◦A is created by taking “shortcuts” from EC. (These are indeed shortcuts, because of the triangle inequality.) So cost(A) 2cost(OPT). 18

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1.5 TSP Approximation Algorithm (Known as Christofides Heuristics) Professor Nicos Christofides extended the 2.0 TSP and published that the worst-case ratio of the extended algorithm was 3/2. 19

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1.5 Approximation Algorithm: 1. Compute MST graph T. 2. Compute a minimum-weighted matching graph M between all Odd- degree vertices. 3. Combine T and M as edge set and Compute an Euler Cycle. 4. Traverse each vertex taking shortcuts to avoid visited nodes. 20

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1.5 Approximation What is a Minimum-weighted Matching? It creates a MWM on a set of the nodes having an odd degree. Why odd degree? Property of Euler Cycle (if all edges all of even-degree, EC exists) Why 1.5 TSP? MST < Euler Cycle <= MWM+MST <= 1.5 TSP (need to show MWM <= ½ MST) 21

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1.5-approximation Claim: The above algorithm gives a 1.5- approximation for the TSP problem (in a metric graph with the triangle inequality). Proof: Denote: OPT, A, and EC as before. M is the set of edges in the minimum matching. 22

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1.5-approximation Proof Continued: cost(M) cost(OPT)/2 ◦The minimum tour on any subset B of vertices has a cost of at most cost(OPT) A Hamiltonian cycle in B can be extracted from OPT by “skipping” vertices not in B, resulting in a tour lighter than OPT, due to the triangle inequality. If B is of even size then divide the minimum tour on B into two disjoint matchings of B. The lighter matching of the two (denote by M’) has weight at most cost(OPT)/2. 23

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1.5-approximation Proof Continued: cost(M) cost(M’) cost(OPT)/2. ◦The number of odd vertices in a graph is even. (Always, but Why?) ◦Choose B to be the set of odd vertices in T. ◦M’ is not necessarily the minimum matching of B, but cost(M) cost(M’). 24

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1.5-approximation Proof Continued: EC is an Euler Cycle in. cost(EC) = cost(T)+cost(M) 1.5cost(OPT). cost(A) cost(EC) 1.5cost(OPT) 25

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26 h c b a d e g f MST h c b d e g f Input (assume Euclidean distances) a Example

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27 h c b a d e g f Walk W a b c b h b a d e f e g e d a h c b a d e g f Hamiltonian Cycle H a b c h d e f g a 2-approx.

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h c b a d e g f MST T h c b e g f Matching M 1.5 approx.

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h c b a d e g f G' = MST + Matching h c b e g f Matching M Matching

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h c b e g f h c b e g f Optimal Tour * on Odd Nodes Separating the optimal tour of B into 2 matchings

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